05a cpt 5 lecture notes f16 - university of...

L38-Wed-30-Nov-2016-Review-Cpt-5-for-Exam-4-HW38 page 71

L38-Wed-30-Nov-2016-Review-Cpt-5-for-Exam-4-HW38 Check out old exam questions on my web site.


NOTE THIS IS g f x NOT f g x

1 1

21

22 1

1 1 1 111 3 3 32 111 11

1 1

x x

x xx

f fx

xx x xx

x x

g

x

g g

x

Find Domain g f x :

2Domain of :1

1 01

f xx

xx

1 :3

3 03

xDomain ofx

xx

So,

:| 1, 3,1 1,3 3,

Domain g f xx x x


Domain f x : x ≠ 1 or ,1 1,

To find range, use the fact that range of f x is domain of 1f x . So, find the inverse by

exchanging x and y and then solving for y:

1

1

2:1

2:1

1 22

22

2

2

xf x yx

yf x xy

x y yxy x y

xy y xy x x

xyx

xf xx

Range f x = Domain 1f x : x ≠ -2 or , 2 2,


If possible, make all the bases the same and then equate the exponents.

•

•

2

2

2

2

2

22 3 1

4 3 1

4 3 1

2

2

19 273

3 3 3

3 3 33 3

Equate exponents:3 4 1

3 4 1 03 1 1 0

3 1 0 1 01 1

3

x x

x x

x x

x x

x xx x

x xx or x

x or x

This problem worked out because we were able to make the bases the same. If we cannot do that, we would have to use substitution and logs:


33 3 33

1 1log log log 3 327 3

OR

3

3

3

1log27

1327133

3 33

y

y

y

y

y

OR

3

1log271 1.43136log 3

27 0.47712log 3


We need to solve

02

xx

The numerator and denominator have zeros at 0 and 2, respectively. So, breakup the number line

into intervals using these values to see when

02

xx

:

Interval: 0 2

Try x -1 1 3

Signs of 2x

x N

N P

N P

P

Sign of f x P N P

0f x ? yes no yes

Note that we cannot include 0 because we cannot take the log of 0.

Note that we cannot include 2 because that would cause a division by 0.

So, the domain is


Convert this to an exponential and solve:

3

1log 38

1812

x

x

x

We could also use change of base to solve this:

0.693

1log 38

1ln8 3

ln1ln8 ln

32.079 ln

30.693 ln

0.5

x

x

x

x

xe x

x


2 7

2 7

8 10 33108

3log 2 78

3log 782

3.3

x

x

x

x

x


Get the logs together and then change to an exponential:

24 4

24 4

24

4

log 9 3 log 3

log 9 log 3 3

9log 33

3 3log

x x

x x

xx

x x

3x

43

3

log 3 3

3 43 64

67

x

xx

x


The solution process is not obvious. Let’s try to write this in a different way using 6 as a base.

1

2 1

2

2

9 36 6

9 6 6 6

9 6 6 6

9 6 6 6

x x

x x

x x

x x

•

•

•

Now, it looks like we can do a u substitution. Let 6xu .

2

2

9 6

6 9 03 3 0

3 0 3 03 3

u u

u uu u

u or uu or u

•

Now, back substitute:

6

3 6

ln3 ln 6

ln3 ln6ln3 1.099 0.613ln6 1.792

x

x

x

u

x

x

05a cpt 5 lecture notes f16 - university of...

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